Lesson 10: Applications — Area, Volume, and Motion

This lesson spends the integration toolkit on the three classical purchases: areas between curves, volumes of solids of revolution, and motion (now in both directions — differentiate position to get velocity, integrate velocity to get position). The mathematics is one idea repeated: slice the quantity into thin pieces you can compute, express the total as an integral, and evaluate. The skill is in the slicing, and in not confusing what is being sliced.

Area between curves: slice vertically

The area between y = f(x) (top) and y = g(x) (bottom) on [a, b] is ∫ from a to b of (f − g) dx — each vertical strip has height (top − bottom) and width dx. The first job is always finding where the curves cross, since the crossing points set the limits and the order of top and bottom can swap there.

Worked Example 1. Area between y = 2x and y = x^2. Intersections: x^2 = 2x gives x = 0 and x = 2. On (0, 2) the line is on top (test x = 1: 2 > 1):

Area = ∫ from 0 to 2 of (2x − x^2) dx = [ x^2 − x^3/3 ] from 0 to 2 = 4 − 8/3 = 4/3

Volumes: discs and washers

Rotate the region under y = f(x) about the x-axis: each cross-section is a disc of radius f(x), so V = π∫ f(x)^2 dx. Rotate the region between two curves and the discs become washers: V = π∫ (R^2 − r^2) dx, outer radius squared minus inner radius squared.

Worked Example 2. Rotate the region under y = sqrt(x) on [0, 4] about the x-axis: V = π∫ from 0 to 4 of (sqrt(x))^2 dx = π∫ from 0 to 4 of x dx = π·8 = 8π ≈ 25.1. (The integrand collapses — (sqrt(x))^2 = x — a small mercy built into many such problems.) Worked Example 3. Rotate the region between y = sqrt(x) (top) and y = x (bottom) on [0, 1] about the x-axis. Outer radius sqrt(x), inner radius x:

V = π∫ from 0 to 1 of ( (sqrt(x))^2 − x^2 ) dx = π∫ from 0 to 1 of (x − x^2) dx = π(1/2 − 1/3) = π/6 ≈ 0.52

Motion, completed

Given v(t) and an initial position, integrate: s(t) = s(0) + ∫ from 0 to t of v. Worked Example 4. v(t) = 6t − 12, s(0) = 5. Then s(t) = 3t^2 − 12t + 5. On [0, 3]: v < 0 until t = 2, then positive. Positions: s(0) = 5, s(2) = 12 − 24 + 5 = −7, s(3) = 27 − 36 + 5 = −4. Displacement = s(3) − s(0) = −9 m; total distance = |−7 − 5| + |−4 − (−7)| = 12 + 3 = 15 m. Displacement integrates v; distance integrates |v| — the same distinction as Lesson 7's particle, now run in reverse from velocity to position.

Average value

The average value of f on [a, b] is (1/(b − a))∫ from a to b of f dx — the height of the rectangle with the same area. Worked Example 5. Temperature T(t) = 20 + 5 sin(π t/12) over a 24-hour day: the sine completes a full period and averages to 0, so the average temperature is exactly 20° — the integral of the sine term over [0, 24] is 0 by symmetry. Averages of periodic functions over whole periods always collapse to the constant term this way.

Application: designing a vessel

A winery wants a tank whose profile is the curve y = sqrt(x) (metres) rotated about the x-axis from x = 0 to x = 4. Worked Example 2 already computed its capacity: 8π ≈ 25.1 m³ ≈ 25,100 litres. The calculus did the engineering: no prototype, no filling experiment. Real vessel design uses exactly this — a profile curve chosen for structural or aesthetic reasons, and a washer/disc integral to certify capacity before anything is built. If the profile changes to y = sqrt(x) + 0.5, the same setup (new integrand, same structure) re-certifies the design in one line.

Common mistakes

• Writing (f − g)^2 in washer problems. Wrong: a washer's area is π(R^2 − r^2). Right: subtract the squares, never square the difference — (R − r)^2 would compute the area of a disc with radius equal to the gap, a different and wrong solid.
• Forgetting π. Volumes of revolution carry π from the circular cross-sections; an answer of 8 instead of 8π for Worked Example 2 is the classic dropped factor.
• Top minus bottom with the roles swapped mid-interval. If the curves cross inside [a, b], split the integral at the crossing. A single (f − g) across a swap produces a signed area that cancels real area away.
• Reporting displacement as distance. In Worked Example 4, −9 and 15 are both "correct integrals" — of different integrands. Read the question: distance means |v|.

Exam Experience

1. The area between y = x^2 and y = x on [0, 1] is

• (A) 0
• (B) 1/6
• (C) 1/3
• (D) 1/2

Answer: (B). On [0, 1], x ≥ x^2, so area = ∫ from 0 to 1 of (x − x^2) dx = 1/2 − 1/3 = 1/6.

2. The region under y = 1/x on [1, e] rotated about the x-axis has volume

• (A) π(1 − 1/e)
• (B) π/e
• (C) π(1 + 1/e)
• (D) 2π(1 − 1/e)

Answer: (A). V = π∫ from 1 to e of (1/x^2) dx = π[ −1/x ] from 1 to e = π(1 − 1/e) ≈ 1.99.

3. A particle with v(t) = 4 − t^2 and s(0) = 0 travels, on [0, 3], a total distance of

• (A) 3
• (B) 16/3
• (C) 23/3
• (D) 25/3

Answer: (C). v changes sign at t = 2. With s(t) = 4t − t^3/3: s(0) = 0, s(2) = 8 − 8/3 = 16/3, s(3) = 12 − 9 = 3. Distance = |16/3 − 0| + |3 − 16/3| = 16/3 + 7/3 = 23/3 ≈ 7.67. Choice (A) is the displacement s(3) − s(0) = 3 — the standard trap.

Free-response question. The region R is bounded by y = sqrt(x), y = 0, and x = 4. (a) Find the area of R. (b) Find the volume when R is rotated about the x-axis. (c) A solid has base R and cross-sections perpendicular to the x-axis that are squares. Find its volume.

Scoring guideline.

• (a) 1 point. ∫ from 0 to 4 of sqrt(x) dx = [ (2/3) x^(3/2) ] from 0 to 4 = (2/3)(8) = 16/3.
• (b) 1 point. V = π∫ from 0 to 4 of x dx = 8π (Worked Example 2).
• (c) 2 points. 1 point for the setup: each square has side sqrt(x), so area x, and V = ∫ from 0 to 4 of x dx. 1 point for the value 8. (The deliberate echo of (b) without π tests whether the solver knows where π comes from — cross-sections here are squares, not circles.)

Going deeper: shells — the second way to slice a volume

Worked Example 3 used washers (slices perpendicular to the axis). Cylindrical shells slice parallel to the axis instead: rotating a vertical strip at position x, of height f(x) and thickness dx, sweeps out a thin tube of volume ≈ 2π·(radius)·(height)·(thickness) = 2π x f(x) dx. For the region under y = sqrt(x) on [0, 4] rotated about the y-axis:

V = 2π ∫ from 0 to 4 of x·sqrt(x) dx = 2π · (2/5) [ x^(5/2) ] from 0 to 4 = 2π · (2/5)(32) = 128π/5 ≈ 80.4

Now the verification that makes this more than a formula: the same volume by washers in y (x = y^2, y from 0 to 2, outer radius 4, inner radius y^2): V = π ∫ from 0 to 2 of (16 − y^4) dy = π (32 − 32/5) = 128π/5 ✓. Two slicings, one solid, one answer — the strongest kind of check applied mathematics offers. The practical rule: choose the slicing whose slices you can actually describe. Regions described as y = f(x) with rotation about the y-axis beg for shells; rotation about the x-axis begs for discs or washers. Fighting the geometry with the wrong slice produces integrals that are correct and unusable, which in practice means wrong.

One more slicing direction: integrate with respect to y

Regions trapped between a parabola and a line often slice more kindly horizontally. The region between x = y^2 and x = y + 2 (curves written as x of y): intersections where y^2 = y + 2, so y = 2 or y = −1; on that interval the line is the right boundary. Horizontal strips have length (right − left) = (y + 2) − y^2, so:

Area = ∫ from −1 to 2 of (y + 2 − y^2) dy = [ y^2/2 + 2y − y^3/3 ] from −1 to 2 = (10/3) − (−7/6) = 9/2

The vertical-slice version requires splitting at the parabola's vertex and solving both curves for y — legal, and miserable. Choosing the variable of integration is a modelling decision: slice perpendicular to whichever axis makes every strip's length a single difference. Examiners set these regions precisely to test whether you look before you integrate.